2 is correct ....................................
![x^ \frac{m}{n}= \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%20%5Cfrac%7Bm%7D%7Bn%7D%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D%20%20)
pemdas, so exponent first before multiply
4(x^1/2)=4x^2
this is different from
(4x)^1/2
so
![x^ \frac{1}{2}= \sqrt[2]{x^1}](https://tex.z-dn.net/?f=x%5E%20%5Cfrac%7B1%7D%7B2%7D%3D%20%5Csqrt%5B2%5D%7Bx%5E1%7D%20%20)
times that by 4
4√x
Supposing all were child tickets:
145x$5=$725
$1153-$725=$428 (diff. between above price and original price)
$9-$5=$4 (diff. between adult price and child price)
$428÷$4=107 (no. of adult tickets)
Ans: 107
Assuming the root sign is over both the numerator and denominator, find the square root of both numbers (6 and 11). The two answers are positive and negative 6/11 because multiplying two negatives makes a positive.
Answer:
Step-by-step explanation:
Given
Required
Solve
From the analysis attached to the question, we can solve this equation by subtracting 5 from both sides:
